Question

A 10eV electron is circulating in a plane at right angle to a uniform field of magnetic infuction 10^-4 Wb/m^2 [ = 1.0 gauss ].The orbital radius of the electron is
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Answer 1

Solution : 11 cm

Explanation

If Charged particle is moving perpendicular to the direction of B, it experiences a maximum force which acts perpendicular to the direction of B as well as v.

Hence, this force will provide the require centripetal force and the charged particle will describe a circular path in the magnetic field if radius r and is given by

mv^2 / r = qvB

Now, KE of electron = 10 eV

=> 1/2 mv^2 = 10 eV

1/2 × ( 9.1 × 10^-31 ) v^2 = 10 × 1.6 × 10^-19

=> v^2 = ( 2 × 10 × 1.6 × 10^-19 ) /9.1 ×10^-31

=> v^2 = 3.52 × 10^12

=> v = 1.88 × 10^6 m

Now, radius of circular path,

r = mv / qB

r = ( 9.1 × 10^-31 × 1.88 × 10^6 ) / ( 1.6 × 10^-19 × 10^-4 )

r = 11cm

The orbital radius of the electron

is 11cm !

Answer 2

Answer:

11 cm

Given : K=10eV B=10

−4

Wb/m

2

Using : mv=Bqr where mv=

2Km

⟹r=

B

2

q

2

2Km

∴ r=

(10

−8

)e

2

2×9.1×10

−31

×10e

=1.06×10

−2

m ≈11 cm